The first assignment for CS480 (Translators) requests that we use Forth as a pocket calculator, rather than teaching the immensely powerful composition strategies for which it’s valued in the real world.
Since our first exposure to the language is a deep dive into the syntax of floating-point computation, no single tutorial on the web answers all the strange assortment of introductory and advanced questions that my classmates and I are running into.
Here’s what I’ve learned so far:
Remember PEMDAS. The quickest way to construct an expression tree for a given mathematical equation is to work from right to left in PEMDAS, then progress downward: The root of the tree is the first subtraction sign in the equation if there is one, otherwise the first addition sign, etc. Everything to the left of the sign which ended up as the root of the tree goes on its left side; everything to the right goes on the right. Then you build each little subtree, following the same rule.
Deep within the introduction of the book Starting Forth, it explains what the comments mean. If you learn nothing else from the introduction, this little section will make lots of other docs make sense:
To communicate stack effects in a visual way, Forth programmers conventionally use a special stack notation in their glossaries or tables of words. We’re introducing the stack notation now so that you’ll have it under your belt when you begin the next chapter.
Here is the basic form:
( before -- after )
The dash separates the things that should be on the stack (before you execute the word) from the things that will be left there afterwards. For example, here’s the stack notation for the word .:
. ( n -- )
(The letter “n” stands for “number.”) This shows that . expects one number on the stack (before) and leaves no number on the stack (after).
Here’s the stack notation for the word +.:
+ ( n1 n2 -- sum )
The floating point stack is separate from the default, or integer, stack. It has different arithmetic operators than the integer stack – typically same as the integer operator, but prefaced with an f.
F@ ( adr --) ( f: -- x) F! ( adr --) ( f: x --) F+ ( f: x y -- x+y) F- ( f: x y -- x-y) F* ( f: x y -- x*y) F/ ( f: x y -- x/y) FEXP ( f: x -- e^x) FLN ( f: x -- ln[x]) FSQRT ( f: x -- x^0.5)
s>f moves the top value of the integer stack onto the floating point stack. For this assignment, it won’t make sense to ever move floating point values back to the integer stack.
Declaring variables is pretty straightforward:
variable a (initialize an intger named a) 23 a ! (assign the value 23 into a) a @ (put a's value onto the integer stack) .s (hey look, the integer stack has 23 on it!)
Remember that you have to preface everything with an f to make it apply to the floating-point stack:
fvariable z (initialize a floating-point variable named z) 69e z f! (z gets the value 69) z f@ (put value of z onto floating-point stack) f.s (Show the floating-point stack.)
When an operator has both an integer and a float as its arguments, when do you put the int onto the float stack? According to a former instructor who used to teach this course, the correct solution is to insert a new node for the s>f conversion between the parent and its integer child in the expression tree, before the post-order traversal. If you try to do the conversion after the traversal and it occurrs at the wrong time, you could end up accidentally swapping the order of the operator’s children. This is harmless for addition and multiplication, but would break everything if it occurred with subtraction or division.
There are two ways to do recursion in gforth. If you say recursive after the name of the word you’re creating, you are able to call the word by name in its definition. Alternately, you can say recurse instead of the name of the word whenever you need to call it from within itself.
Since both fibonacci and factorial are used in the homework, I’ll use a function that sums from 0 to the argument as an example. In Python, it’d look like:
def sum(i): if (i == 0): return 0 return i + sum(i-1)
Using forth’s recursive command, it’ll look like:
: sum recursive dup ( top copy of the argument will get destroyed by comparison ) 0= ( take argument off the stack, replace it with true or false ) if ( we'll only take this branch if the argument was == 0 ) exit else ( we have i on the stack, need to return i + the sum ) dup ( extra copy, keep an i for adding later ) 1- ( decrement, now stack has i-1 on top to be the arg to next call ) sum ( recursively call with that i-1 we made ) + ( yay postfix! ) endif ;
If we don’t use recursive, it looks almost identical except for the actual recursive call:
: othersum dup 0= if exit else dup 1- recurse ( does exactly the same thing as the call to sum ) + endif ;
So testing them out will look something like this:
3 sum . 6 ok 4 sum . 10 ok 3 othersum . 6 ok 4 othersum . 10 ok
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